The brightness depends on power dissipation. For the same voltage, DC provides more power to the bulb than AC (where the RMS value is lower than the peak value), so the bulb shines brighter with DC.

The RMS value of a sinusoidal current is given by \( I_{rms} = \frac{I_0}{\sqrt{2}} \), where \( I_0 \) is the peak current. Here, \( I_0 = 4 \) A, so \( I_{rms} = \frac{4}{\sqrt{2}} \) A.

The RMS (root mean square) value of an AC voltage is related to its peak value by \( V_{rms} = \frac{V_0}{\sqrt{2}} \), where \( V_0 \) is the peak voltage.

The power dissipated is given by \( P = V_{rms} \times I_{rms} \times \cos\phi \), where ϕ is the phase difference. Here, ϕ = 0 (both are sine functions with same phase), so \( \cos\phi = 1 \). \( V_{rms} = \frac{5}{\sqrt{2}} \), \( I_{rms} = \frac{2}{\sqrt{2}} \), so \( P = \frac{5}{\sqrt{2}} \times \frac{2}{\sqrt{2}} \times 1 = \frac{10}{2} = 5 \) W.

The power factor \( \cos\phi = \frac{R}{Z} \). Given \( \phi = \frac{\pi}{3} \), \( \cos(\pi/3) = 0.5 \). So \( Z = \frac{R}{\cos\phi} = \frac{10}{0.5} = 20 \) Ω.

Effective voltage (RMS voltage) = Peak voltage / √2 = 423 / 1.414 ≈ 300 volts.

Resonant frequency \( f_0 = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{0.1 \times 100 \times 10^{-6}}} = \frac{1}{2\pi \times 10^{-2}}} = \frac{100}{2\pi} \approx 50 \) Hz.

Since voltage and current are in phase, average power \( P = V_{rms} \times I_{rms} = \frac{200}{\sqrt{2}} \times \frac{5}{\sqrt{2}} = \frac{1000}{2} = 500 \) W. However, looking at the options, there seems to be a discrepancy. The correct calculation gives 500W, but the closest option is 100W. This suggests there may be an error in the question or options.

Time to reach from zero to peak = T/4 = 1/(4f) = 1/(4×50) = 5 × 10⁻³ sec. Peak current = RMS × √2 = 10 × 1.414 ≈ 14.14 A.

In a purely inductive circuit, the voltage leads the current by π/2 radians (90°). In a capacitive circuit, current leads voltage by π/2 radians.

Comparing with standard form V = V₀ sin(2πft), we see 2πf = 120π ⇒ f = 60 Hz. RMS voltage = V₀/√2 = 240/1.414 ≈ 170 V, but this doesn't match the options. The question might have an error in options.

RMS current = V/R = 200/40 = 5 A. Peak current = RMS × √2 ≈ 5 × 1.414 ≈ 7 A. However, the closest option is 5.0 A, suggesting the question might be asking for RMS current.

At resonance in an RLC circuit, the inductive and capacitive reactances cancel each other, making the circuit purely resistive. Therefore, current and voltage are in phase, impedance is minimum (equal to R), current is maximum, and power factor is unity.

Peak voltage = RMS voltage × √2 = 220 × 1.414 ≈ 311 V.

The Q-factor of a series resonant circuit can be expressed as \( Q = \frac{1}{R}\sqrt{\frac{L}{C}} \) or \( Q = \frac{\omega L}{R} \) at resonance frequency. Both expressions are equivalent at resonance.

Power dissipated \( P = I_{rms}^2 R \). Since \( I_{rms} = \frac{I_0}{\sqrt{2}} \), \( P = \left(\frac{I_0}{\sqrt{2}}\right)^2 R = \frac{I_0^2}{2} R \).

The standard frequency of AC mains in India is 50 Hz (cycles per second).

Power \( P = V_{rms} I_{rms} \cos\phi \), where ϕ is the phase difference. Here ϕ = π/2, so cos(π/2) = 0, making P = 0. This represents a purely inductive or capacitive circuit where no real power is dissipated.

Power factor \( \cos\phi = R/Z \). Given \( \cos\phi = 0.8 \) and Z = 10Ω, so R = Z × cosϕ = 10 × 0.8 = 8Ω.

For a sinusoidal AC waveform, the RMS value is \( \frac{1}{\sqrt{2}} \) times the peak value.

In a purely capacitive AC circuit, the current leads the voltage by 90° (π/2 radians). This is because in a capacitor, current is proportional to the rate of change of voltage.

Power \( P = V_{rms} I_{rms} \cos\phi \). Here \( V_{rms} = \frac{100}{\sqrt{2}} \), \( I_{rms} = \frac{100}{\sqrt{2}} \), and ϕ = π/3 ⇒ cos(π/3) = 0.5. So \( P = \frac{100}{\sqrt{2}} \times \frac{100}{\sqrt{2}} \times 0.5 = \frac{10000}{2} \times 0.5 = 2500 \) W. This doesn't match the options, suggesting an error in the question or options.

Inductive reactance \( X_L = 2\pi f L \). If frequency is halved from 100 Hz to 50 Hz, the reactance is also halved from 20 Ω to 10 Ω.

\( \omega = 2\pi f = 2\pi \times 50 = 100\pi \) rad/s. At \( t = \frac{1}{600} \) s, \( \omega t = 100\pi \times \frac{1}{600} = \frac{\pi}{6} \) rad. So \( E = 10 \sin(\pi/6) = 10 \times 0.5 = 5 \) V.

Domestic AC voltage is always quoted as the RMS (root mean square) value, which is 220 V in this case. The peak voltage would be higher (about 311 V).

Comparing with \( I = I_0 \sin(2\pi ft) \), we see \( 2\pi f = 200\pi \) ⇒ f = 100 Hz. Time period T = 1/f = 1/100 sec. Time to reach first peak = T/4 = 1/400 sec.

RMS current = Peak current / √2 = 10 / √2.

For a symmetrical AC waveform (like sine wave), the average value over one complete cycle is zero because the positive and negative halves cancel each other out.

RMS voltage = 141 / √2 ≈ 100 V. Angular frequency ω = 628 = 2πf ⇒ f = 628/(2π) ≈ 100 Hz.

Capacitive reactance \( X_C = \frac{1}{2\pi f C} \). If frequency increases 4 times (from 100 Hz to 400 Hz), reactance decreases 4 times (from 20 Ω to 5 Ω).

A DC ammeter measures average current. For AC, the average over a complete cycle is zero, so the DC ammeter would show zero deflection.

Peak voltage = RMS voltage × √2 = 220 × 1.414 ≈ 311 V.

Assuming 10V is the RMS value, peak value = RMS × √2 = \( 10 \sqrt{2} \) V.

Power factor \( \cos\phi \) is maximum (equal to 1) when ϕ = 0, which occurs in a purely resistive circuit. In circuits with reactance, the power factor is less than 1.

For a sinusoidal AC current, the RMS value is related to the peak value by \( I_{rms} = \frac{I_0}{\sqrt{2}} \).

At resonance in a series LCR circuit, the inductive and capacitive reactances cancel each other, leaving only the resistance. Therefore, the impedance is minimum and equal to the resistance.

The resistance (real part of impedance) of a coil remains the same for DC and AC. However, the total impedance in AC includes both resistance and reactance.

Hot wire ammeters measure RMS current. Peak current = RMS × √2 = 10 × 1.414 ≈ 14.14 A.

The process of converting AC to DC is called rectification, typically done using diodes.

In an RLC circuit, current becomes maximum at the resonant frequency when \( X_L = X_C \), making the impedance purely resistive and minimum.

Power in an AC circuit is given by \( P = V_{rms} I_{rms} \cos\phi \), where ϕ is the phase difference between voltage and current. Thus, power depends on the phase relationship.

Angular frequency ω = 377 = 2πf ⇒ f = 377/(2π) ≈ 60 Hz.

Adding capacitance to an inductive circuit can cancel out some of the inductive reactance, reducing the phase angle and improving the power factor (bringing it closer to 1).

RMS voltage = Peak voltage / √2 = 707 / 1.414 ≈ 500 V.

In a pure inductor, the current lags the voltage by 90°, making the power factor zero (cos90° = 0). Therefore, the average power consumed is zero, though there is energy storage and release in the magnetic field.

The phase difference is π/2 (90°), so power factor cos(π/2) = 0. Therefore, average power \( P = V_{rms} I_{rms} \cos\phi = 0 \).

RMS voltage = Peak voltage / √2 = 120 / 1.414 ≈ 84.8 V.

The ratio of peak value to RMS value for a sinusoidal AC waveform is √2 (approximately 1.414).

The bandwidth is defined as the range between the half-power points (where current is 1/√2 ≈ 70.7% of maximum current).

The current has a +π/3 phase term, meaning it leads the voltage by π/3 radians (60°). This is characteristic of a circuit with capacitive reactance dominating.